Regular hexagon $ABCDEF$ is divided into six smaller equilateral triangles, such as $\triangle ABG$, shown in boldface in the diagram.  By connecting every other vertex, we obtain a larger equilateral triangle $\triangle ACE$, also shown in boldface.  Compute the ratio $[\triangle ABG]/[\triangle ACE]$. [asy]
size(150); defaultpen(linewidth(0.8)); dotfactor=5;
pair[] hex = new pair[6];
string[] hexlabels = {"$C$","$B$","$A$","$F$","$E$","$D$"};
hexlabels.cyclic=true;
hex[0] = dir(0);
for(int i = 1; i <= 6; ++i){

hex[i] = dir(60*i);

draw(hex[i] -- hex[i-1]);

dot(hexlabels[i],hex[i],hex[i]);
}
draw(hex[0]--hex[3]); draw(hex[1]--hex[4]); draw(hex[2]--hex[5]);
draw(hex[0]--hex[2]--hex[4]--cycle,linewidth(1.3));
draw(hex[1]--hex[2]--(0,0)--cycle,linewidth(1.3));
dot("$G$",(0,0),2*S);
[/asy]
Answer: Each small congruent right triangle in the diagram has the same area, which we will call $K$.  Since $\triangle ABG$ consists of two small triangles, $[\triangle ABG]= 2K$.  Similarly, $\triangle ACE$ is built from six small triangles, so $[\triangle ACE] = 6K$.  Hence the ratio of these areas is $2K/6K = \boxed{\frac{1}{3}}$.